Le Chatelier's Principle - Chromate-dichromate - C12!4!07 (1) 2o4759

Le Chatelier’s Principle: Investigating the Chromate-Dichromate Equilibrium C12-4-07 Introduction: Most of the chemical reactions you have observed in the laboratory seem to have gone to completion – that is, all the reactants appear to have been used up to form the products. Actually, all chemical reactions are reversible, even though the extent of reversibility may sometimes be very small. In many reactions that you will encounter in the laboratory, you must concern yourself with their reversibility and the possibilities of controlling the extent of reversibility. In this experiment you will study some reactions in which there is appreciable reversibility and in which the presence of the reactants and products can be readily observed by noting colour changes or the formation of a precipitate. In an aqueous solution the chromate ion, CrO42- (aq), can be converted to the dichromate ion, Cr2O72- (aq), and, conversely, the dichromate ion, Cr2O72- (aq) ion can be converted to the chromate ion, CrO42- (aq) ion. The extent to which these reactions occur is dependent upon the concentration of the hydrogen ion, H+ (aq), in the solution. The H+(aq) concentration can be increased by adding a source of H+(aq) - hydrochloric acid, HCl. The H+ (aq) concentration can be decreased by adding a solution of sodium hydroxide, NaOH, which contains the hydroxide ion, OH- (aq). The OH- (aq) reacts with H+ (aq) to form H2O, causing a shift in the equilibrium. The net-ionic equilibrium reactions of potassium chromate and potassium dichromate, respectively, are:

K2CrO4

chromate-dichromate (K2CrO4) equilibrium: 2CrO42- + 2H+

↔ H2O + Cr2O72- (eqn. 1)

*Note that there is predominantly CrO42- ions present with a small amount of Cr2O72- ions.

K2Cr2O7

dichromate-chromate (K2Cr2O7) equilibrium: Cr2O72- + 2OH- ↔ H2O + 2CrO42- (eqn. 2) *Note that there is predominantly Cr2O72- ions present with a small amount of CrO42- ions.

Le Chatelier’s Principle states that “If a stress is applied to a reaction at equilibrium, the reaction will shift to offset the stress applied”. The addition of hydrogen ions or hydroxide ions constitutes a stress, as does the removal of either ion, and so it can be expected that the above equilibria will shift in response to concentration changes in these ions. For example, in eqn. 1, if [H+] is increased by adding HCl, there are more ions available for reaction and the reaction will favour the formation of products. Conversely, if [H+] is decreased (by adding OH-) there are fewer ions available for the forward reaction and the reaction shifts reverse to favor reactant formation. 1

Materials: 1 Spot plate 10 eyedroppers 0.1 M potassium chromate, K2CrO4 0.1 M potassium dichromate, K2Cr2O7 1 M sodium hydroxide, NaOH 1 M hydrochloric acid, HCl

1 M acetic acid, CH3COOH 1 M nitric acid, HNO3 1 M potassium hydroxide, KOH 1 M ethanol, C2H5OH 1 M aqueous ammonia, NH3 0.1 M barium nitrate, Ba(NO3)2

Procedure: Part A – The Chromate-Dichromate and The Dichromate-Chromate Equilibrium: *Record observations in the Data Table below: 1. A 0.1 M potassium chromate, K2CrO4, and a 0.1 M potassium dichromate, K2Cr2O7, solution will serve as sources for the ions, CrO42- (aq) and Cr2O72- (aq). Record the colour of each solution. 2. Place 10 drops (about ½ ml) of each solution (K2CrO4 and K2Cr2O7) into 2 separate wells in a spot plate. Add 1 M sodium hydroxide, NaOH, alternately to each well one drop at a time. Record the colour changes observed. Retain these wells for step 5. What inferences can you make based on any changes observed? 3. Repeat the procedure of step 2 with fresh solutions in clean wells, but add 1 M hydrochloric acid, HCl, drop by drop alternately to each well. Record the colour changes observed. Retain these wells for step 4. Again what inferences can you make based on any changes observed? 4. Add 1 M NaOH, drop be drop, to one of the wells obtained in step 3 until a change is noted. Again what inferences can you make based any change observed? 5. Add 1 M HCl, drop by drop, to one of the wells obtained in step 2 until a change is noted.

Data Part A K2CrO4

K2Cr2O7

Step 1 (record color) Step 2 (add NaOH) Step 3 (add HCl) Step 4 (add NaOH to step 3) Step 5 (add HCl to step 2)

Part A Analysis: 1.

(a)

Explain how the chromate-dichromate equilibrium (eqn. 1) shifts in response to an increase or decrease in H+ ions. Express this both in words and as a chemical equation.

(b)

Explain how the chromate-dichromate equilibrium (eqn. 1) shifts in response to an increase or decrease in OH- ions. Express this both in words and as a chemical equation.

(c)

Use Le Chatelier’s Principle to explain the colour changes noted in Part A, steps 3 and 5. Ensure you refer to what is happening in of why changes from one ion to another occur as a result of chemical species available.

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2.

(a)

Explain how the dichromate-chromate equilibrium (eqn. 2) shifts in response to an increase or decrease in H+ ions. Express this in words and as an equation.

(b)

Explain how the dichromate-chromate equilibrium (eqn. 2) shifts in response to an increase or decrease in OH- ions. Express this in words and as an equation.

(c)

Use Le Chatelier’s Principle to explain the colour changes noted in Part A, steps 2 and 4.

Part B – Additional Experiments on the Chromate-Dichromate and the Dichromate-Chromate Equilibrium: *Record observations in the Data Table below: 1. In the following investigation you are going to be adding various acids and bases to the chromate and dichromate solutions. Again consider what effect acids (which donate H+) and bases (which donate OH-) have on these ions. Repeat the procedure used in Part A, steps 2 and 3, using the solutions of K2CrO4 and K2Cr2O7 but instead of using HCl and NaOH, testing each with 4 to 5 drops of 1 M solutions of each of the following: a. b. c. d. e.

acetic acid, CH3COOH nitric acid, HNO3 potassium hydroxide, KOH ethanol, C2H5OH aqueous ammonia, NH3

Data Part B: K2CrO4

K2Cr2O7

CH3COOH (Acid) HNO3 KOH (Base) C2H5OH NH3 (aq)

Part B Analysis: 1. (b)

(a)

Which substances in Part B caused the colour to change from that of the Cr2O72- (aq) ion to that of the CrO42- (aq) ion?

Which substances in Part B caused the colour to change from that of the CrO42- (aq) ion to that of the Cr2O72- (aq) ion? 2.

(a)

What ions do the solutions you listed in 1(a) have in common? (b)

3.

What ions do the solutions you listed in 1(b) have in common?

Give an explanation for the results noted when: (a) ethanol, C2H5OH, was added (b) the solution of aqueous ammonia, NH3 (aq), was added. 4.

On the basis of your answers to # 2, predict some additional substances which might have the same effect on the chromate-dichromate or dichromate-chromate equilibrium as those categorized in #1(a) and (b).

5.

Explain in words how the addition of acids or bases affect the chromate-dichromate equilibrium. 3

Part C – The Equilibrium between Barium Chromate, BaCrO4, and Barium Dichromate, BaCr2O7. 1. Place 10 drops of 0.1 M K2CrO4 in a clean well. Add 2 drops of 1 M NaOH. Add 0.1 M barium nitrate, Ba(NO3)2, one drop at a time, until a change is noted. Record the result. Retain this well for step 4. The reaction between K2CrO4 and Ba(NO3)2 is: K2CrO4

+

Ba(NO3)2 ↔

BaCrO4 + 2KNO3

What was the purpose of adding NaOH prior to the reaction? Explain the result in of Le Chatelier’s principle of adding the barium nitrate on the equilibrium. 2. Place 10 drops of 0.1 M K2Cr2O7 in a clean well. Add 2 drops of 1 M HCl, then 10 drops of 0.1 M Ba(NO3)2. Record the result. Retain this well and solution for step 5. The reaction between K2Cr2O7 and Ba(NO3)2 is: K2Cr2O7

+

Ba(NO3)2 ↔

BaCr2O7 + 2KNO3

What was the purpose of adding HCl prior to the reaction? 3. Record your conclusions about the relative solubilities of BaCrO4 and BaCr2O7 from your observations in steps 1 and 2. Confirm your results using a solubility chart. 4. To the solution saved from step 1, add 1 M HCl, drop by drop, until a change is noted. Record your observations. 5. Again explain why this has happened in of Le Chatelier’s principle. 6. To the solution saved from step 2, add 1 M NaOH, drop by drop, until a change is noted. Record your observations. 7. Again explain why this has happened in of Le Chatelier’s principle 8. Suggest a way to reverse the changes and reactions you observed in steps 4 and 5. Try these experiments. 9. Place 10 drops of 0.1 M K2Cr2O7 in a clean well and the same amount of 0.1 M K2CrO4 in another clean well. Add a few drops of 0.1 M Ba(NO3)2 to each. Record your observations. 10. Again, explain why this has happened in of Le Chatelier’s principle

Analysis Part C: 1.

From your observations in step 9, what can you conclude about the relative equilibrium concentrations of CrO42- (aq) ion in each of the solutions 0.1 M K2Cr2O7 and 0.1 M K2CrO4 before the Ba(NO3)2 solution is added? 4

Summary: Le Chatelier's Principle states that if a system is at equilibrium and something is changed so that it is no longer at equilibrium, the system will respond in an effort to counteract that change. If more reactant is added, the equilibrium will shift forward in order to consume some of the extra reactant (since there are more ions available for reaction), resulting in more product. If some of the product is removed from the system, the equilibrium will shift forward to produce more of that product. The equilibrium can be shifted reverse by either adding product to or removing reactant from the system. Yellow chromate ion and orange dichromate ion are in equilibrium with each other in aqueous solution. The more acidic the solution, the more the equilibrium is shifted to favour the dichromate ion. As nitric acid is added to the potassium chromate solution, the yellow colour turns to orange. When sodium hydroxide is added to the potassium chromate solution, the orange colour turns back to yellow. The sodium hydroxide reacts with hydrogen ions, removing them from the solution. When one reactant is removed from an equilibrium system, the equilibrium shifts reverse, in this case forming the yellow chromate ion again. Barium reacts with chromate ions (but not dichromate ions) to form an insoluble salt. Solid barium chromate can be converted to soluble barium dichromate by adding a source of H+ ions to the solution. This reaction can be reversed again by adding OH- ions, which decreases the H+ concentration, thus favouring barium chromate.

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